The area bounded between the curve $x^2=y$ and the line $y=4 x$ is
- $\frac{32}{3}$ sq. units
- $\frac{8}{3}$ sq. units
- $\frac{1}{3}$ sq. units
- $\frac{16}{3}$sq. units
Solution
Required area is shaded.
Point of intersection of given curves are $(0,0)$ and $(4,16)$
$\begin{aligned}
& \therefore A=\int_0^4\left(4 x-x^2\right) d x \\
& =4 \int_0^4 x d x-\int_0^4 x^2 d x=4\left[\frac{x^2}{2}\right]_0^4-\left[\frac{x^3}{3}\right]_0^4 \\
& =2(16)-\frac{(4)^2}{3}=32-\frac{64}{3}=\frac{32}{3} \text { sq. units }
\end{aligned}$Asked in: MHT CET 2021 (22 Sep Shift 1)